I am new to boolean algebra so having these two question to simplify hope you guys will help me,

Q- Using Demorgan's theorem, show that

a. (A + B)' (A' + B')' = 0
b. A + A'B + A'B' = 1

 

hi,
as far as part b is concerned here is the proof:

A+ A'B +A'B' = A + A' ( B + B' ) = A + A' = 1

because by demorgan laws, A + A' =1

part a is not clear what is that emoticon doing in between? B)

 

Thanks for answering b while in a it is B and parenthesis instead of emotion.

Originally posted by arvind@Nov 9 2004, 05:29 AM
hi,
as far as part b is concerned here is the proof:

A+ A'B +A'B' = A + A' ( B + B' ) = A + A' = 1

because by demorgan laws, A + A' =1

part a is not clear what is that emoticon doing in between? B)

[post=3479]Quoted post[/post]​

 

DeMorgans theorem says:
(x AND y)' = x' OR y'
(x OR y)' = x' AND y'


For your example a:
a. (A + B )' (A' + B')' = 0

This is composed of two expressions ANDed together
Exp1 is (A + B )'
Exp2 is (A' + B')'

Convert each expression using DeMorgans Theorem
Exp1 becomes A'B'
Exp2 becomes AB

So, does ( A'B' )( AB ) = 0 ?

( A'B' )( AB ) = A'B'AB -> associative property
= A'AB'B -> associative
= (A'A)(B'B') -> associative
= (0)(0) -> complements identity
= 0 -> Done

So, (A + B )' (A' + B')' = 0

Perion

 

well sikander , one of the best ways to solve boolean algebra probs is to use Venn diagrams , u can really solve problems very quickly .


|--------|--|------|
|--------|--|------|
|----1-- |3|---2--|
|--------|- |-------|
1 is A\B
2 is B\A
3 is A intersection B

a+a'b=the total area which is a+b
now a' wud include all the space including B
b' all space include all space including A
their intersection wud include all the space except a and B
so a+b + all space not including a+b=all space = 1

well ,the figure looks complicated but this is really easy .